library(pracma)to_LaTeX <- function(A){
rows <- apply(A, MARGIN=1, paste, collapse = " & ")
matrix_string <- paste(rows, collapse = " \\\\ ")
return(paste("\\begin{bmatrix}", matrix_string, "\\end{bmatrix}"))
}
get_properties <- function(A){
m = nrow(A)
n = ncol(A)
if (m != n){
square = FALSE
}
else{
square = TRUE
}
equality <- c()
if (square){
for (i in 1:m){
for (j in 1:n){
if (A[i, j] == A[j, i]){
append(equality, TRUE)
}else{
append(equality, FALSE)
}
}
}
symmetric <- isTRUE(all(equality == TRUE))
}else{
symmetric <- FALSE
}
t_A = t(A)
t_A_m = nrow(t_A)
t_A_n = ncol(t_A)
properties <- list("A" = A, "A_rows" = m, "A_cols" = n, "square" =
square, "symmetric" = symmetric, "AT" = t_A,
"AT_rows" = t_A_m, "AT_cols" = t_A_n)
}
count_zero_rows <- function(A){
properties <- get_properties(A)
copy <- properties$A
m <- properties$A_rows
zero_row_count <- 0
for (r in 1:m){
s <- sum(copy[r, ])
if (s == 0){
zero_row_count <- zero_row_count + 1
}else{
next
}
}
zero_row_count
}
move_all_zero_rows <- function(A){
properties <- get_properties(A)
copy <- properties$A
m <- properties$A_rows
n <- properties$A_cols
zero_row_count <- 0
checking <- TRUE
while (checking){
row_sums <- c()
for (r in 1:(m - 1)){
s <- sum(copy[r, ])
if (s == 0){
copy <- copy[-r, ]
zero_row_count <- zero_row_count + 1
break
}else{
next
}
row_sums <- append(row_sums, s)
}
if (all(row_sums > 0)){
checking <- FALSE
}
}
if (zero_row_count > 0){
for (z in 1:zero_row_count){
copy <- rbind(copy, rep(0, n))
}
}
result <- list("copy" = copy, "zero_row_count" = zero_row_count)
}
perform_gauss_jordan_elimination <- function(A){
properties <- get_properties(A)
copy <- properties$A
m <- properties$A_rows
n <- properties$A_cols
result <- move_all_zero_rows(copy)
copy <- result$copy
col <- 1
row <- 0
while (col < (n + 1) & (row + 1) <= m){
if (sum(copy[(row + 1):m, col]) == 0){
col <- col + 1
}else{
ind <- 0
for (i in (row + 1):m){
if (copy[i, col] != 0){
ind <- i
break
}
}
row <- row + 1
#swap row_i and row_r
row_i <- copy[ind, ]
row_r <- copy[row, ]
copy[ind, ] <- row_r
copy[row, ] <- row_i
#make leading 1
scalar <- 1 / copy[row, col]
copy[row, ] <- copy[row, ] * scalar
#make 0 entries
for (k in 1:m){
if (copy[k, col] != 0 & k != row){
scalar <- copy[k, col] / copy[row, col]
copy[k, ] <- -1 * scalar * copy[row, ] + copy[k, ]
}
}
col <- col + 1
}
}
copy
}What is the rank of the matrix \(A\)?
A = rbind(c(1, 2, 3, 4),
c(-1, 0, 1, 3),
c(0, 1, -2, 1),
c(5, 4, -2, -3))The rank of the matrix \(A\) is 4, which the function below determines by counting the non-zero rows in the reduced row-echelon form of the matrix \(A\).
compute_rank <- function(A){
properties <- get_properties(A)
copy <- properties$A
m <- properties$A_rows
print("Original matrix: ")
print(copy)
luA <- lu(copy)
REF <- luA$U
print("Reduced row-echelon form: ")
print(REF)
zero_row_count <- count_zero_rows(REF)
rank <- m - zero_row_count
print("Rank: ")
print(rank)
rank
}
A_rank <- compute_rank(A)## [1] "Original matrix: "
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3
## [1] "Reduced row-echelon form: "
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4.000
## [2,] 0 2 4 7.000
## [3,] 0 0 -4 -2.500
## [4,] 0 0 0 1.125
## [1] "Rank: "
## [1] 4Given an \(mxn\) matrix where \(m > n\), what can be the maximum rank? The minimum rank, assuming that the matrix is non-zero?
The maximum rank for a matrix is the lesser of the number of rows and columns, so here the maximum rank would be \(n\). The minimum rank, since we’re assuming the matrix has at least one non-zero element, would be 1.
What is the rank of the matrix \(B\)?
B = rbind(c(1, 2, 1),
c(3, 6, 3),
c(2, 4, 2))The rank of the matrix \(B\) is 1. The compute_rank function I used previously for the matrix \(A\) only works if a matrix can be LU factorized. The matrix \(B\) cannot be LU factorized because two of its leading minors, the determinants for its second and third submatrices, are both 0. It can still be put in reduced row-echelon form though, so an alternate rank computing function is defined below to account for that.
compute_rank_when_LU_factorization_fails <- function(A){
properties <- get_properties(A)
copy <- properties$A
m <- properties$A_rows
print("Original matrix: ")
print(copy)
REF <- perform_gauss_jordan_elimination(copy)
print("Reduced row-echelon form: ")
print(REF)
zero_row_count <- count_zero_rows(REF)
rank <- m - zero_row_count
print("Rank: ")
print(rank)
rank
}
B_rank <- compute_rank_when_LU_factorization_fails(B)## [1] "Original matrix: "
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2
## [1] "Reduced row-echelon form: "
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0
## [1] "Rank: "
## [1] 1Compute the eigenvalues and eigenvectors of the matrix \(A\). You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.
A = rbind(c(1, 2, 3),
c(0, 4, 5),
c(0, 0, 6))
I <- matrix(0, nrow = 3, ncol = 3)
diag(I) <- 1
A_print <- to_LaTeX(A)
I_print <- to_LaTeX(I)Finding the characteristic polynomial:
\(Det(\begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix} - \lambda\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}) = 0\)
B = rbind(c("1 - \U03BB", "2", "3"),
c("0", "4 - \U03BB", "5"),
c("0", "0", "6 - \U03BB"))
B_print <- to_LaTeX(B)\(Det(\begin{bmatrix} 1 - λ & 2 & 3 \\ 0 & 4 - λ & 5 \\ 0 & 0 & 6 - λ \end{bmatrix}) = 0\)
characteristic_polynomial_step1 <- "((1 - \U03BB) * (((4 - \U03BB) * (6 - \U03BB)) - ((0))) - 0 + 0))"
characteristic_polynomial_step2 <- "(1 - \U03BB) * (24 - 4\U03BB - 6\U03BB + \U03BB^2)"
characteristic_polynomial_step3 <- "(1 - \U03BB) * (24 - 10\U03BB + \U03BB^2)"
characteristic_polynomial_step4 <- "24 - 10\U03BB + \U03BB^2 - 24\U03BB + 10\U03BB^2 - \U03BB^3"
characteristic_polynomial_step5 <- "-\U03BB^3 + 11\U03BB^2 - 34\U03BB + 24"\(((1 - λ) * (((4 - λ) * (6 - λ)) - ((0))) - 0 + 0)) = 0\)
\((1 - λ) * (24 - 4λ - 6λ + λ^2) = 0\)
\((1 - λ) * (24 - 10λ + λ^2) = 0\)
\(24 - 10λ + λ^2 - 24λ + 10λ^2 - λ^3 = 0\)
\(-λ^3 + 11λ^2 - 34λ + 24 = 0\)
Finding the roots of the characteristic polynomial, which are the eigenvalues:
coefs <- c(24, -34, 11, -1)
roots <- Re(polyroot(coefs))Eigenvalues: \([1, 4, 6]\)
Eigenvectors:
Finding the eigenvector for \(\lambda = 1\):
B_char = rbind(c("1 - 1", "2", "3"),
c("0", "4 - 1", "5"),
c("0", "0", "6 - 1"))
B_num = rbind(c(0, 2, 3),
c(0, 3, 5),
c(0, 0, 5))
B_char_print <- to_LaTeX(B_char)
B_num_print <- to_LaTeX(B_num)
var <- as.matrix(rbind("x", "y", "z"))
z <- matrix(0, nrow = 3, ncol = 1)
var_print <- to_LaTeX(var)
z_print <- to_LaTeX(z)
eq1 <- "2y + 3z = 0"
eq2 <- "3y + 5z = 0"
eq3 <- "5z = 0"
sol <- "x = 1, y = z = 0"
evec <- as.matrix(rbind(1, 0, 0))
evec_print <- to_LaTeX(evec)\(\begin{bmatrix} 1 - 1 & 2 & 3 \\ 0 & 4 - 1 & 5 \\ 0 & 0 & 6 - 1 \end{bmatrix} = \begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix}\)
\(\begin{bmatrix} 0 & 2 & 3 \\ 0 & 3 & 5 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
\(2y + 3z = 0\)
\(3y + 5z = 0\)
\(5z = 0\)
\(x = 1, y = z = 0\)
Eigenvector: \(\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}\)
Finding the eigenvector for \(\lambda = 4\):
B_char = rbind(c("1 - 4", "2", "3"),
c("0", "4 - 4", "5"),
c("0", "0", "6 - 4"))
B_num = rbind(c(-3, 2, 3),
c(0, 0, 5),
c(0, 0, 2))
B_char_print <- to_LaTeX(B_char)
B_num_print <- to_LaTeX(B_num)
eq1 <- "-3x + 2y + 3z = 0"
eq2 <- "5z = 0"
eq3 <- "2z = 0"
sol <- "x = 2/3, y = 1, z = 0"
evec <- as.matrix(rbind("2/3", "1", "0"))
evec_print <- to_LaTeX(evec)\(\begin{bmatrix} 1 - 4 & 2 & 3 \\ 0 & 4 - 4 & 5 \\ 0 & 0 & 6 - 4 \end{bmatrix} = \begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix}\)
\(\begin{bmatrix} -3 & 2 & 3 \\ 0 & 0 & 5 \\ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
\(-3x + 2y + 3z = 0\)
\(5z = 0\)
\(2z = 0\)
\(x = 2/3, y = 1, z = 0\)
Eigenvector: \(\begin{bmatrix} 2/3 \\ 1 \\ 0 \end{bmatrix}\)
Finding the eigenvector for \(\lambda = 6\):
B_char = rbind(c("1 - 6", "2", "3"),
c("0", "4 - 6", "5"),
c("0", "0", "6 - 6"))
B_num = rbind(c(-5, 2, 3),
c(0, -2, 5),
c(0, 0, 0))
B_char_print <- to_LaTeX(B_char)
B_num_print <- to_LaTeX(B_num)
eq1 <- "-5x + 2y + 3z = 0"
eq2 <- "-2y + 5z = 0"
sol <- "x = 8/5, y = 5/2, z = 1"
evec <- as.matrix(rbind("8/5", "5/2", "1"))
evec_print <- to_LaTeX(evec)\(\begin{bmatrix} 1 - 6 & 2 & 3 \\ 0 & 4 - 6 & 5 \\ 0 & 0 & 6 - 6 \end{bmatrix} = \begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix}\)
\(\begin{bmatrix} -5 & 2 & 3 \\ 0 & -2 & 5 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}\)
\(-5x + 2y + 3z = 0\)
\(-2y + 5z = 0\)
\(x = 8/5, y = 5/2, z = 1\)
Eigenvector: \(\begin{bmatrix} 8/5 \\ 5/2 \\ 1 \end{bmatrix}\)